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3 years ago

1. Why would it be difficult to recycle polyvinyl chloride into plastic bags?

Chemistry

2 answers:

mario62 [17]3 years ago

8 0

Answer:

1. Recycling polyvinyl chloride (PVC) into plastic bags would be difficult because PVC is made of a harder material that normal plastic bags are made of. GL with the portfolio my man.

Explanation:

Georgia [21]3 years ago

6 0

Answer:

Explanation:

1. It is difficult to recycle polyvinyl chloride into plastic bags because it is a very hard plastic. Polyvinyl chloride is a widely used plastic because it is durable and stable with an average life span of 30 years.

2. HDPE is a good material for boats because of its high tensile strength and low density. HDPE stands for High-density polyethylene. It is lightweight and impervious to damage.

3. HDPE. This is because it is unaffected by alkaline solution. It's corrosive nature gives it resistance to damage.

4. HDPE is used in making plastic lumber.

It is also used in producing geomembrane.

5. To separate the mixture of PETE and HDPE, you would need to employ the flotation method of separating mixtures. Mixtures are separated based on their hydrophobic and hydrophilic abilities. The mixtures are placed in a condensed tube, and stirred. The substance with a lighter weight stays at the top while the heavier one stays below. They are then removed separately.

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What is the reactant(s) in the chemical equation below?

butalik [34]

Answer:

C. 3CO(g) + Fe2O3(s)

Explanation:

The substance(s) to the hath left of the arrow in a chemical equation art hath called reactants. A reactant is a substance yond is presenteth at the starteth of a chemical reaction. The substance(s) to the right of the arrow art hath called products. A product is a substance yond is presenteth at the endeth of a chemical reaction

So in this example, 3CO(g) + Fe2O3(s) art the reactants.

The 2Fe(S) + 3CO2(G) art the products.

Desire I holp! Has't a most wondrous day!

Hope I helped! Have a great day!

7 0

3 years ago

The volume of a pond being studied for the effects of acid rain is 35 kiloliters (kL). There are 1,000 liters (L) in 1 kL and 1

Lerok [7]

Answer:

35,000,000,000 mL

Explanation:

You first multiply 35 times 1000.

35,000 L

Now you multiply 35,000 times 10^6

35,000,000,000 mL

4 0

3 years ago

In order to reduce the exposure to organic solvents like turpentine, some art instructors recommend the students clean brushes a

kykrilka [37]

Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.

To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.

6 0

3 years ago

30 POINTS!!! Please Help With this!!!!

Goshia [24]

X is dependent Y is independent.... number of students is Y flavor is X

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3 years ago

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0

3 years ago